Find the equation of the tangent line at the point where the curve crosses itself. x = t^3-6t, y = t^2.

Here is the solution for this specific problem:

Based from the graph, the curve will intersect itself at the y-axis, i.e. x = 0. 

t^3 - 6t = 0 
t(t^2 - 6) = 0 
t = 0 or t = ± √6 

dx/dt = 3t^2 - 6 
dy/dt = 2t 

dy/dx = 2t/(3t^2 - 6) 

@ t = 0, dy/dx = 0. 
x = 0, y = 0 

y = 0 

@ t = √6, dy/dx = 2√6/12 = √6/6 
x = 0, y = 6 

y - 6 = (√6/6) x 
y = (√6/6)x + 6 

@ t = -√6, dy/dx = -2√6/12 = -√6/6 
x = 0, y = 6 

y - 6 = (-√6/6) x 
y = (-√6/6)x + 6

So the equations of the tangent line at the point where the curve crosses itself are: y = (√6/6)x + 6 and y = (-√6/6)x + 6. I am hoping that these answers have satisfied your queries and it will be able to help you in your endeavors, and if you would like, feel free to ask another question.

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step-by-step explanation:

fa tha first one you gonna add y to both sides

it look like this

x-y+y=6+y

x=y+6

therefore your answer is

x =y+6

fa tha second one

you gonna flip tha equation

it gonna look like x-4y=y

then add 4 to both sides

it gonna look like tha

x-4 +4 =y +4

so therefore tha answer is x=y+4

step-by-step explanation: